Xét \(5-x=0\Leftrightarrow x=5\)

\(x-1=0\Leftrightarrow x=1\)

\(2+3x=0\Leftrightarrow x=-\dfrac{2}{3}\)

Bảng xét dấu:

Để VT\(\le\)0 <=>\(\left[{}\begin{matrix}-\dfrac{2}{3}\le x\le1\\x\ge5\end{matrix}\right.\)

Vậy...

(5-x)(x-1)(2+3x) ≤ 0

↔ 5-x≤0 <=> x≥5 (1)

x-1 ≤ 0<=> x≤ 1 (2)

2+3x ≤ 0 <=> x≤ -2/3 (3)

Từ (1),(2),(3) ta có:

  x≥5 or x≤1 or x≤ -2/3

chúc bạn học tốt !!!

\(2+\dfrac{3\left(x+1\right)}{3}\le3-\dfrac{x-1}{4}\)

\(\Leftrightarrow2+x+1\le\dfrac{12}{4}-\dfrac{x-1}{4}\)

\(\Leftrightarrow x+3\le\dfrac{13-x}{4}\)

\(\Leftrightarrow\dfrac{4x+12}{4}\le\dfrac{13-x}{4}\)

\(\Leftrightarrow4x+12\le13-x\)

\(\Leftrightarrow4x+x\le13-12\)

\(\Leftrightarrow5x\le1\)

\(\Leftrightarrow x\le\dfrac{1}{5}\)

Vậy: \(x\le\dfrac{1}{5}\) 

\(2+\dfrac{3\left(x+1\right)}{3}\le3-\dfrac{x-1}{4}\)

\(\Leftrightarrow\dfrac{12x+36}{12}\le\dfrac{33-3x}{12}\)

\(\Leftrightarrow12x+36\le33-3x\)

\(\Leftrightarrow12x+3x\le-36+33\)

\(\Leftrightarrow15x\le-3\)

\(\Leftrightarrow x\le\dfrac{-1}{5}\)

\(a,\dfrac{2x-1}{3}< \dfrac{x+6}{2}\)

\(\Leftrightarrow\dfrac{4x-2}{6}< \dfrac{3x+18}{6}\)

\(\Leftrightarrow4x-2< 3x+18\)

\(\Leftrightarrow4x-3x< 2+18\)

\(\Leftrightarrow x< 20\)

\(b,\dfrac{5\left(x-1\right)}{6}-1>\dfrac{2\left(x+1\right)}{3}\)

\(\Leftrightarrow\dfrac{5x-11}{6}>\dfrac{4x+4}{6}\)

\(\Leftrightarrow5x-11>4x+4\)

\(\Leftrightarrow5x-4x>11+4\)

\(\Leftrightarrow x>15\)

ĐKXĐ: \(x>0\)

\(3\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< 2\left(x+\dfrac{1}{4x}+1\right)-9\)

\(\Leftrightarrow3\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< 2\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)^2-9\)

Đặt \(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=a>0\)

\(\Rightarrow3a< 2a^2-9\Rightarrow2a^2-3a-9>0\)

\(\Rightarrow\left(a-3\right)\left(2a+3\right)>0\)

\(\Rightarrow a-3>0\Rightarrow a>3\)

\(\Rightarrow\sqrt{x}+\dfrac{1}{2\sqrt{x}}>3\Leftrightarrow2x+1>6\sqrt{x}\)

\(\Leftrightarrow2x-6\sqrt{x}+1>0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}>\dfrac{3+\sqrt{7}}{2}\\0\le\sqrt{x}< \dfrac{3-\sqrt{7}}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x>\dfrac{8+3\sqrt{7}}{2}\\0\le x< \dfrac{8-3\sqrt{7}}{2}\end{matrix}\right.\)

Điều kiện xác định : \(1\le x\le7\)

Bất phương trình chuyển thành :

\(x-1+2\sqrt{7-x}-2\sqrt{x-1}-\sqrt{\left(x-1\right)\left(7-x\right)}\le0\)

Đặt \(a=\sqrt{x-1};b=\sqrt{7-x}\) ta có :

\(a^2-2a-ab+2b\le0\)

\(\Leftrightarrow\left(a-b\right)\left(a-2\right)\le0\)

\(\Leftrightarrow\left[{}\begin{matrix}a\le b\\a\le2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1\le7-x\\x-1\le4\end{matrix}\right.\)

Sau đó tìm x

- Với \(-1\le x\le0\Rightarrow\left\{{}\begin{matrix}VT>0\\VP\le0\end{matrix}\right.\) BPT hiển nhiên đúng

- Với \(0< x< 1\) hai vế đều dương, bình phương:

\(\Leftrightarrow1+\sqrt{1-x^2}\ge x^2\left(1+4\left(1-x^2\right)+4\sqrt{1-x^2}\right)\)

\(\Leftrightarrow1-x^2+\sqrt{1-x^2}-4x^2\left(1-x^2\right)-4x^2\sqrt{1-x^2}\ge0\)

\(\Leftrightarrow\left(1-4x^2\right)\left(1-x^2\right)+\sqrt{1-x^2}\left(1-4x^2\right)\ge0\)

\(\Leftrightarrow\left(1-4x^2\right)\left(1-x^2+\sqrt{1-x^2}\right)\ge0\)

\(\Leftrightarrow1-4x^2\ge0\) (do ngoặc sau luôn dương)

\(\Rightarrow0< x\le\frac{1}{2}\)

Vậy nghiệm của BPT đã cho là \(-1\le x\le\frac{1}{2}\)

Có sai sót chút xíu, sửa lại, bạn để ý nhé

\(\frac{x+1}{x-2}\ge2\)

\(\Rightarrow x+1\ge2\left(x-2\right)\)

\(\Leftrightarrow x+1\ge2x-4\)

\(\Leftrightarrow-x\ge-5\)

\(\Leftrightarrow x\le5\)

vậy tập nghiệm của bất phương trình là: {x| x\(\le\)5}

ĐKXĐ: x khác 2 

\(\Leftrightarrow x+1\ge2\text{x}-2\)

\(\Leftrightarrow x-3\le0\)

\(\Leftrightarrow x\le3\)

1.a)|−7x|=3x+16

Vì |-7x| ≥ 0  nên 3x+16 ≥ 0 ⇔ x ≥ \(\dfrac{-16}{3}\)    (*)

Với đk (*), ta có: |-7x|=3x+16

\(\left[\begin{array}{} -7x=3x+16\\ -7x=-3x-16 \end{array} \right.\) ⇔  \(\left[\begin{array}{} -7x-3x=16\\ -7x+3x=-16 \end{array} \right.\)

⇔ \(\left[\begin{array}{} x=-1,6 (t/m)\\ x= 4 (t/m) \end{array} \right.\)

b) \(\dfrac{x-1}{x+2}\) - \(\dfrac{x}{x-2}\) = \(\dfrac{5x-8}{x^2-4}\)

⇔ \(\dfrac{(x-1)(x-2)}{x^2-4}\) - \(\dfrac{x(x+2)}{x^2-4}\) = \(\dfrac{5x-8}{x^2-4}\)

⇒ x² - 2x - x + 2 - x² - 2x = 5x - 8  

⇔ -5x - 5x = -8 - 2

⇔ -10x = -10

⇔ x=1

2.7x+5 < 3x−11

⇔ 7x - 3x < -11 - 5

⇔ 4x < -16

⇔ x < -4

bạn tự biểu diễn trên trục số nha !